to find the zeros or roots or solutions, we simply set f(x) = 0, now let's keep in mind that even multiplicity zeros, do not cross the x-axis, they simply bounce off it, whilst odd multiplicity zeros do cross the x-axis.
[tex] \bf \stackrel{f(x)}{0}=\left( x+\frac{1}{4} \right)^2(x-1)^5\\\\-------------------------------\\\\0=\left( x+\frac{1}{4} \right)^{\stackrel{\stackrel{\downarrow }{even}}{2}}\implies 0=x+\cfrac{1}{4}\implies -\cfrac{1}{4}=x\impliedby \textit{even multiplicity}\\\\-------------------------------\\\\0=(x-1)^{\stackrel{\stackrel{\downarrow }{odd}}{5}}\implies 0=x-1\implies 1=x\impliedby \textit{odd multiplicity} [/tex]
